Problem: Complete the square to solve for $x$. $x^{2}-4x-21 = 0$
Begin by moving the constant term to the right side of the equation. $x^2 - 4x = 21$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-4$ , half of it would be $-2$ , and squaring it gives us ${4}$ $x^2 - 4x { + 4} = 21 { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x - 2 )^2 = 25$ Take the square root of both sides. $x - 2 = \pm5$ Isolate $x$ to find the solution(s). $x = 2\pm5$ So the solutions are: $x = 7 \text{ or } x = -3$ We already found the completed square: $( x - 2 )^2 = 25$